My thought

In fact, it's natural to find the general term expression of $a_n$. Indeed, $a_n$ could be expressed explicitly as follows $$a_n=x^{2^{n-1}}+\frac{1}{x^{2^{n-1}}}~,$$ where $x$ is any of the two roots of the equation $x^2-3x+1=0$, i.e. $x=\dfrac{3 \pm \sqrt{5}}{2},$ and $n=1,2,\cdots.$

It's simple to prove that by mathematical induction. Let $n=1$, then $$x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}=x+\dfrac{1}{x}=\dfrac{x^2+1}{x}=\dfrac{3x}{x}=3=a_1.$$Assume that $a_k=x^{2^{k-1}}+\dfrac{1}{x^{2^{k-1}}},$ then $$a_{k+1}=a_k^2-2=\left(x^{2^{k-1}}+\dfrac{1}{x^{2^{k-1}}}\right)^2-2=x^k+\frac{1}{x^k}+2-2=x^k+\frac{1}{k}.$$ Thus, the general term formula above of $a_n$ holds for any $n=1,2,\cdots.$ From this, we have have $$a_1=\frac{1+x^2}{x},a_2=\frac{1+x^4}{x^2},\cdots，a_n=\frac{1+x^{2^n}}{x^{2^{n-1}}}.$$ But can those offer some help to solve the problem？Besides,the general term expression is what someone told me, but he didn't tell me how to get this. Who can show me?

Another thought

Let $b_n=\dfrac{a_n-\sqrt{a_n^2-4}}{2}$, where $a_n>2$. Then \begin{align*}b_{n+1}&=\dfrac{a_{n+1}-\sqrt{a_{n+1}^2-4}}{2}=\dfrac{a_{n}^2-2-\sqrt{(a_{n}^2-2)^2-4}}{2}=\dfrac{a_{n}(a_n-\sqrt{a_n^2-4})-2}{2}=a_nb_{n}-1.&\end{align*}
Hence, $$b_n=\frac{1}{a_n}+\frac{b_{n+1}}{a_n}.$$Therefore,$$b_1=\frac{1}{a_1}+\frac{b_2}{a_1}=\frac{1}{a_1}+\frac{1}{a_1}\left(\frac{1}{a_2}+\frac{b_3}{a_2}\right)=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{b_3}{a_1a_2}=\cdots$$
Expand like this repeatedly till infinitely. Then we have $$b_1=\dfrac{1}{a_1}+\dfrac{1}{a_1a_2}+\dfrac{1}{a_1 a_2 a_3}+\cdots.$$ It follows that $$\dfrac{1}{a_1}+\dfrac{1}{a_1a_2}+\dfrac{1}{a_1 a_2 a_3}+\cdots=b_1=\dfrac{3-\sqrt{5}}{2}.$$

$\begingroup$@mengdie1982: If you have a solution, you should provide it. Otherwise, how are we supposed to know if we have "a different one"? Besides, we don't want to waste time telling you things you already know.$\endgroup$
– BlueMay 11 '18 at 10:17

1 Answer
1

Notice that $a_1=3>2$. Suppose that $a_k>2$, then $a_{k+1}=a_k^2-2>2^2-2=2$. By mathematical induction, we have $a_n>2,$ further, $a_n^2>a_n+2$. Hence, $a_{n+1}=a_n^2-2>(a_n+2)-2=a_n.$ This shows that $a_n$ is an increasing integer sequence. Therefore, $a_n \to +\infty$ as $n \to \infty$.
Thus, we have $$\lim_{n \to \infty}\frac{\dfrac{1}{a_1 a_2 \cdots a_{n+1}}}{\dfrac{1}{a_1 a_2 \cdots a_{n}}}=\lim_{n \to \infty}\dfrac{1}{a_{n+1}}=0.$$ According to the ratio test of positive series, we have that $\sum\limits_{n=1}^{\infty}\dfrac{1}{a_1a_2\cdots a_n}$ is convergent. Hence, let it be $\dfrac{a_1-\lambda}{2},$ where $\lambda$ is an unknown real number we need to find.
It's simple to find that $$\frac{a_2-\lambda a_1}{2}=\frac{1}{a_2}+\frac{1}{a_2a_3}+\cdots$$$$\frac{a_3-\lambda a_1a_2}{2}=\frac{1}{a_3}+\frac{1}{a_3a_4}+\cdots$$$$\vdots$$$$\frac{a_n-\lambda a_1a_2\cdots a_{n-1}}{2}=\frac{1}{a_n}+\frac{1}{a_na_{n+1}}+\cdots$$
Since $$0<\frac{1}{a_n}+\frac{1}{a_na_{n+1}}+\cdots <\frac{1}{a_n}+\frac{1}{a_n^2}+\cdots=\frac{1}{a_n-1}，$$ hence $$0<\frac{a_n-\lambda a_1a_2\cdots a_{n-1}}{2}<\frac{1}{a_n-1}.$$
According to the Squeeze Theorem,$$\lim_{n \to \infty}\frac{a_n-\lambda a_1a_2\cdots a_{n-1}}{2}=0.$$ Thus, $$\lambda=\lim_{n \to \infty}\frac{a_n}{a_1a_2\cdots a_{n-1}}.$$
Moreover, from $a_{n+1}=a_n^2-2$, we have $$a_n^2-4=a_{n-1}^2(a_{n-2}^2-4)=\cdots=a_1^2a_2^2\cdots a_{n-1}^2(a_1^2-4).$$
Thus, $$\left(\frac{a_n}{a_1a_2\cdots a_{n-1}}\right)^2=\frac{4}{a_1^2a_2^2\cdots a_{n-1}^2}+(a_1^2-4).$$
Taking the limits as $n \to \infty$ of the both sides, we have $$ \lambda^2=0+5=5.$$ Then,$$\lambda=\sqrt{5}.$$ As a result, $$\frac{1}{a_1}+\frac{1}{a_1 a_2}+\frac{1}{a_1a_2a_3}+\cdots+\frac{1}{a_1a_2 \cdots a_n}+\cdots=\frac{3-\sqrt{5}}{2}.$$ This is exactly what we want to prove.